Let $A,B$ be the points on the coordinate plane with coordinates $(t-4,-1)$ and $(-2,t+3)$, respectively. The square of the distance between the midpoint of $\overline{AB}$ and an endpoint of $\overline{AB}$ is equal to $t^2/2$. What is the value of $t$?
Explanation: The distance between the midpoint of $\overline{AB}$ and an endpoint of $\overline{AB}$ is equal to half of the length of $\overline{AB}$. By the distance formula,

\begin{align*}
AB &= \sqrt{((t-4)-(-2))^2 + ((-1)-(t+3))^2}\\
&= \sqrt{(t-2)^2+(t+4)^2} \\
&= \sqrt{2t^2 + 4t + 20}
\end{align*}Also, we know that $(AB/2)^2 = t^2/2 \Longrightarrow AB = 2\sqrt{t^2/2} = \sqrt{2t^2}$. Setting these two expressions equal and squaring, we obtain $$AB^2 = 2t^2 = 2t^2 + 4t + 20 \Longrightarrow 4t + 20 = 0.$$Thus, $t = \boxed{-5}$.